Question

How do you factor the trinomial `49x^2 + 56x +9`?

Answer 1

Nothing pretty; I got
`color(white)("XXX")(x+(28+sqrt(343))/49)(x+(28-sqrt(343))/49)`

Explanation:

You might check the question; if the last term were `(-9)` instead of `(+9)` then there would be some nice integer terms.

Working with what is given, my only approach would be to apply the quadratic formula:
`color(white)("XXX")x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`
to the general equation form: `ax^2+bx+c`
to get the factors `(x-x_1)(x-x_2)`

In this case
`color(white)("XXX")x_(1,2) = (-56+-sqrt(56^2-4*49*9))/(2*49)`

`color(white)("XXXXXX")=(-56+-sqrt(3136-1764))/(2*49)`

`color(white)("XXXXXX")=(-56+-sqrt(1372))/(2*49)`

`color(white)("XXXXXX")=(-56+-2sqrt(343))/(2*49)`

`color(white)("XXXXXX")=(-28+-sqrt(343))/49`

So the factors are
`color(white)("XXX")(x-((-28-sqrt(343))/49))*(x-((-28+sqrt(343))/49))`

...all this assumes that I didn't make any arithmetic errors.